Problem Description
Steve has an integer array a of length n (1-based). He assigned all the elements as zero at the beginning. After that, he made m operations, each of which is to update an interval of a with some value. You need to figure out ⨁ni=1(i⋅ai) after all his operations are finished, where ⨁ means the bitwise exclusive-OR operator.
In order to avoid huge input data, these operations are encrypted through some particular approach.
There are three unsigned 32-bit integers X,Y and Z which have initial values given by the input. A random number generator function is described as following, where ∧ means the bitwise exclusive-OR operator, << means the bitwise left shift operator and >> means the bitwise right shift operator. Note that function would change the values of X,Y and Z after calling.
Let the i-th result value of calling the above function as fi (i=1,2,⋯,3m). The i-th operation of Steve is to update aj as vi if
where
Input
The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains five space-separated integers n,m,X,Y and Z.
1≤T≤100, 1≤n≤105, 1≤m≤5⋅106, 0≤X,Y,Z<230.
It is guaranteed that the sum of n in all the test cases does not exceed 106 and the sum of m in all the test cases does not exceed 5⋅107.
Output
For each test case, output the answer in one line.
Sample Input
4
1 10 100 1000 10000
10 100 1000 10000 100000
100 1000 10000 100000 1000000
1000 10000 100000 1000000 10000000
Sample Output
1031463378
1446334207
351511856
47320301347
Hint
In the first sample, a = [1031463378] after all the operations.
In the second sample, a = [1036205629, 1064909195, 1044643689, 1062944339, 1062944339, 1062944339, 1062944339, 1057472915, 1057472915, 1030626924] after all the operations.
Source
2018 Multi-University Training Contest 5
没什么好说的，就搞一个最小值然后加一个重要的剪枝，在update里面如果这个区间的最小值已经比加的值大了，就return
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
const int mod=(1<<30);
const int maxn=100005;
typedef unsigned ll;
ll f[15000000], x,y,z;
int n,m;
unsigned rng()
{
unsigned t;
x^=x<<11;
x^=x>>4;
x^=x<<5;
x^=x>>14;
t=x;
x=y;
y=z;
z=x^y^t;
return z;
}
struct node{
int minn,lazy;
}e[maxn<<2];
void pushdown(int rt){
if(e[rt].lazy){
if(e[lson].lazy<e[rt].lazy)
e[lson].lazy=e[rt].lazy;
if( e[rson].lazy<e[rt].lazy)
e[rson].lazy=e[rt].lazy;
if( e[lson].minn<e[rt].lazy)
e[lson].minn=e[rt].lazy;
if( e[rson].minn<e[rt].lazy)
e[rson].minn=e[rt].lazy;
e[rt].lazy=0;
}
}
void update(int nowl,int nowr,int l,int r,int rt,ll aim){
if(e[rt].minn>=aim)return ;
if(nowl>=l&&nowr<=r){
if(e[rt].minn>=aim){
return ;
}
else
{
e[rt].minn=aim;
e[rt].lazy=aim;
return ;
}
}
pushdown(rt);
int mid=(nowl+nowr)/2;
if(mid>=l) update(nowl,mid,l,r,lson,aim);
if(mid<r) update(mid+1,nowr,l,r,rson,aim);
e[rt].minn=min(e[lson].minn,e[rson].minn);
}
ll query(int nowl,int nowr,int l,int r,int rt){
if(nowl>=l&&nowr<=r){
return e[rt].minn;
}
pushdown(rt);
int mid=(nowl+nowr)/2;
if(mid>=l) return query(nowl,mid,l,r,lson);
if(mid<r) return query(mid+1,nowr,l,r,rson);
}
int main(){
int t;
cin>>t;
while(t--){
scanf("%d%d%u%u%u",&n,&m,&x,&y,&z);
memset(e,0,sizeof(e));
for(int i=1;i<=3*m;i++)
f[i]=rng();
for(int i=1;i<=m;i++){
int l=min(f[3*i-2]%n+1,f[3*i-1]%n+1);
int r=max(f[3*i-2]%n+1,f[3*i-1]%n+1);
ll aim=f[3*i]%mod;
update(1,n,l,r,1,aim);
}
unsigned long long ans=0;
for(int i=1;i<=n;i++){
unsigned long long nows=query(1,n,i,i,1);
ans^=(i*nows);
}
cout<<ans<<endl;
}
}